The strong law gives $T_n/n\to 1/\lambda$. Since $T_{N_t}\le t<T_{N_t+1}$ and $N_t\to \infty$, sandwiching yields $N_t/t\to \lambda$ almost surely.

Apply part (a) to the stopping time $N_t+1$ to obtain $E{T}_{N_t+1}={\lambda }^{-1}E(N_t+1)$.

This is Wald's identity. Expand the stopped sum and use $\{M\ge i\}$, which depends only on earlier interarrivals and is independent of ${\xi }_i$: $E{\sum }_{i=1}^M{\xi }_i={\sum }_{i}E[1_{M\ge i}]E{\xi }_i={\lambda }^{-1}EM$.

On disjoint events $A_n$ with $P(A_n)=c/n^2$, set $X_n=n{1}_{A_n}$. Then $X_n\to 0$ almost surely and $E{X}_n^2=c$, but any common dominator has integral at least ${\sum }_{n}nP(A_n)=c\sum 1/n=\infty$. Thus no integrable dominating variable exists.

Yes. The $L^2$ bound makes the family uniformly integrable in $L^1$: $E[|X_n|1_{|X_n|>K}]\le E[X_n^2]/K\le 1/K$. Together with convergence in probability, Vitali's theorem gives $E|X_n|\to 0$.

For every $ϵ>0$, $1_{|X_n-X|>ϵ}\to 0$ almost surely and is dominated by 1, so its expectation, the corresponding probability, tends to zero.

Fatou says $\int \mathrm{liminf}{f}_n\le \mathrm{liminf}\int f_n$ for nonnegative functions. Apply it to $g+f_n$ and $g-f_n$ when $|f_n|\le g\in L^1$ to obtain both inequalities and hence dominated convergence.

The one-dimensional kernel of $d{K}_x$ is the tangent to its level curve; the metric gives two unit normals and $d{K}_x(e_x)>0$ selects one. Isometries preserve the metric, $K$, and this orientation choice, so they carry $e_x$ to $e_{\varphi (x)}$. Therefore $g(x)=d{K}_x(e_x)=|\nabla K(x)|$ is isometry invariant.

An isometry preserves both foliations. Independence makes their level curves transverse near $x$, so their pair of values gives local coordinates. The isometry fixes the intersection corresponding to every pair of nearby values, hence fixes every point in a sufficiently small neighbourhood.

Gauss's Theorema Egregium expresses Gaussian curvature intrinsically from the first fundamental form, hence $K\circ \varphi =K$. Merely preserving curvature level sets is not sufficient: when $K$ is constant, every diffeomorphism does so, though most are not isometries.

If $f\circ \varphi =f$, then $x$ and $\varphi (x)$ have the same value, so every level set is preserved.

An isometry is a smooth diffeomorphism with $I_x(v,w)=I_{\varphi (x)}(d{\varphi }_{x}v,d{\varphi }_{x}w)$.

Not for an arbitrary signed $L^1$ kernel. One can place very tall, very narrow spikes of $g$ at separated locations, with summable masses and total integral one, and matching still narrower spikes of an $L^1$ function accumulating at a Lebesgue point. Along a selected subsequence of dilations, a kernel spike lands on the matching function spike and contributes a fixed amount, while the total widths make the point a Lebesgue point. Thus compact support (or a suitable integrable radial majorant) cannot simply be dropped.

At a Lebesgue point, $f\ast g_n(x)-f(x)=\int g(u)[f(x-u/n)-f(x)]du$. Compact support and boundedness of $g$ reduce this to a constant times the mean oscillation over a ball of radius $O(1/n)$, which tends to zero.

Approximate $f$ in $L^1$ by continuous $g$. The limsup local oscillation of $f$ is bounded by the maximal function of $f-g$ plus the vanishing oscillation of $g$. The weak $(1,1)$ Hardy-Littlewood maximal inequality makes the exceptional set have measure at most $C\Vert f-g{\Vert }_1/ϵ$; let the approximation error tend to zero.

A point $x$ is a Lebesgue point of $f$ when $|B_r{|}^{-1}{\int }_{B_r(x)}|f(y)-f(x)|dy\to 0$ as $r\downarrow 0$.

Approximate $h$ uniformly by finite sums $\sum f_j(x)g_j(y)$; their integral operators have finite-dimensional range and converge in operator norm, so $T_h$ is compact. If $h(x,y)=0$ for $x\le y$, iterated integration over an ordered simplex gives $\Vert T_h^n\Vert \le \Vert h{\Vert }_{\infty }^n/n!$. Thus $r(T_h)=0$, and compact spectral theory gives $\sigma (T_h)=\{0\}$.

$\Vert T_{h}f{\Vert }_{\infty }\le \Vert h{\Vert }_{\infty }\Vert f{\Vert }_{\infty }$, so $T_h$ is bounded, and $h\mapsto T_h$ is linear with operator norm at most one.

Complex Stone-Weierstrass requires a self-conjugate unital subalgebra separating points; apply real Stone-Weierstrass to real and imaginary parts. Finite sums $f(x)g(y)$ form such a self-conjugate unital subalgebra of $C(K\times L)$ and separate points, so they are dense.

$\sigma (T)=\{\lambda :T-\lambda I\text{ is not invertible}\}$ and $r(T)={\sup }_{\lambda \in \sigma (T)}|\lambda |$, well-defined because the spectrum is nonempty compact. If $|\lambda |>\Vert T^n{\Vert }^{1/n}$, factor $T^n-{\lambda }^{n}I$ and use a Neumann series; spectral mapping then excludes $\lambda$. Thus $r(T)\le {\inf }_n\Vert T^n{\Vert }^{1/n}$.